Question: The red parabola shown is the graph of the equation $x = ay^2 + by + c$.  Find $a+b+c$.

[asy]
size(150);
real ticklen=3;
real tickspace=2;

real ticklength=0.1cm;
real axisarrowsize=0.14cm;
pen axispen=black+1.3bp;
real vectorarrowsize=0.2cm;
real tickdown=-0.5;
real tickdownlength=-0.15inch;
real tickdownbase=0.3;
real wholetickdown=tickdown;
void rr_cartesian_axes(real xleft, real xright, real ybottom, real ytop, real xstep=1, real ystep=1, bool

useticks=false, bool complexplane=false, bool usegrid=true) {

import graph;

real i;

if(complexplane) {

label("$\textnormal{Re}$",(xright,0),SE);

label("$\textnormal{Im}$",(0,ytop),NW);

} else {

label("$x$",(xright+0.4,-0.5));

label("$y$",(-0.5,ytop+0.2));

}

ylimits(ybottom,ytop);

xlimits( xleft, xright);

real[] TicksArrx,TicksArry;

for(i=xleft+xstep; i<xright; i+=xstep) {

if(abs(i) >0.1) {

TicksArrx.push(i);

}

}

for(i=ybottom+ystep; i<ytop; i+=ystep) {

if(abs(i) >0.1) {

TicksArry.push(i);

}

}

if(usegrid) {

xaxis(BottomTop(extend=false), Ticks("%", TicksArrx ,pTick=gray

(0.22),extend=true),p=invisible);//,above=true);

yaxis(LeftRight(extend=false),Ticks("%", TicksArry ,pTick=gray(0.22),extend=true),

p=invisible);//,Arrows);

}

if(useticks) {

xequals(0, ymin=ybottom, ymax=ytop, p=axispen, Ticks("%",TicksArry ,

pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize));

yequals(0, xmin=xleft, xmax=xright, p=axispen, Ticks("%",TicksArrx ,

pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize));

} else {

xequals(0, ymin=ybottom, ymax=ytop, p=axispen, above=true, Arrows(size=axisarrowsize));

yequals(0, xmin=xleft, xmax=xright, p=axispen, above=true, Arrows(size=axisarrowsize));

}
};
real lowerx, upperx, lowery, uppery;
real f(real x) {return -(x+4)*(x+4)/2+5;}
lowery = -9;
uppery = 1;
rr_cartesian_axes(-8,7,lowery,uppery);
draw(reflect((0,0),(1,1))*(graph(f,lowery,uppery,operator ..)), red);
[/asy]

Each tick mark on the graph is one unit.
The vertex of the parabola is $(5,-4)$, so the equation of the parabola is of the form \[x = a(y + 4)^2 + 5.\] The parabola passes through the point $(3,-2)$.  Substituting these values into the equation above, we get \[3 = a(-2 + 4)^2 + 5.\] Solving for $a$, we find $a = -1/2$.  Hence, the equation of the parabola is given by \[x = -\frac{1}{2} (y + 4)^2 + 5 = -\frac{1}{2} (y^2 + 8y + 16) + 5 = -\frac{1}{2} y^2 - 4y - 3.\] The answer is $-1/2 - 4 - 3 = \boxed{-\frac{15}{2}}$.